Finite Unions of Closed Sets

Why Finiteness Matters

The previous tutorial considered intersections of closed sets, where any number of closed constraints can be imposed at once. Unions behave differently: the union of finitely many closed sets is closed, but an arbitrary union need not be. This distinction is one of the basic patterns of real-line topology.

The finite-union result was established in the tutorial “Sequential Characterization of Closed Sets.” We will use that theorem by name rather than prove it again. Our aim is to understand what finiteness contributes, how finite unions interact with closure, and how to track limit points when a set is split into finitely many pieces.

Theorem (Finite Unions of Closed Sets, recalled): If $F_1,\ldots,F_r$ are closed subsets of $\mathbb{R}$, where $r$ is a positive integer, then $\bigcup_{j=1}^{r}F_j$ is closed.

The theorem has no requirement that the closed sets be disjoint, bounded, or nonempty. Some of the sets may even equal $\mathbb{R}$, while others may be empty. What matters is that the list has finitely many members. That condition will matter again when we examine limit points: a sequence whose terms lie in a finite union must have infinitely many terms in at least one member of the union.

Using the Finite-Union Theorem

A useful first step is to express a set as a finite union of familiar closed sets. Closed intervals and closed rays are standard examples. Once each piece is known to be closed, the finite-union theorem gives the closedness of the whole set immediately. In many problems the main work is therefore identifying an appropriate finite decomposition.

Worked Example: A Union of Two Closed Intervals

Consider

F=[-4,-1]\cup[2,6].

Both $[-4,-1]$ and $[2,6]$ are closed intervals, so each is closed in $\mathbb{R}$. The Finite Unions of Closed Sets Theorem therefore implies that $F$ is closed. The intervals are separated, but separation is not needed for the theorem; the conclusion would hold even if they overlapped or shared an endpoint.

For example, if the second interval were $[-1,6]$, the union would be $[-4,6]$, still a finite union of closed sets and therefore closed. The theorem is about the operation of taking a finite union, not about the geometric arrangement of its pieces.

Worked Example: A Closed Set Described by a Distance Bound

Let

G=\{x\in\mathbb{R}:|x+2|\geq 1\}.

The inequality $|x+2|\geq1$ holds exactly when $x+2\leq-1$ or $x+2\geq1$. Subtracting $2$ in each inequality gives $x\leq-3$ or $x\geq-1$. Hence

G=(-\infty,-3]\cup[-1,\infty).

Each ray is closed, so their union is closed by the finite-union theorem. The equality also displays why the boundary points are included: $|-3+2|=1$ and $|-1+2|=1$, so both satisfy the defining inequality. Replacing $\geq$ by $>$ would remove those endpoints and give open rays instead, which would be a different set.

Closure Distributes Over Finite Unions

The closure of a set is its smallest closed superset, as established in the previous tutorial. For a finite union, the closure can be found by closing each piece separately and then taking their union. This is a new consequence of the finite-union theorem: it converts a closure calculation for a combined set into separate closure calculations for its components.

Theorem (Closure of a Finite Union): For subsets $A_1,\ldots,A_r$ of $\mathbb{R}$, where $r$ is a positive integer, $$ \overline{\bigcup_{j=1}^{r}A_j} = \bigcup_{j=1}^{r}\overline{A_j}. $$

Proof. Write $A=\bigcup_{j=1}^{r}A_j$. Each $\overline{A_j}$ is closed. By the Finite Unions of Closed Sets Theorem, $\bigcup_{j=1}^{r}\overline{A_j}$ is closed. It contains $A$, because $A_j\subseteq\overline{A_j}$ for every $j$. Since $\overline{A}$ is the smallest closed set containing $A$, it follows that \[ \overline{A}\subseteq\bigcup_{j=1}^{r}\overline{A_j}. \] For the reverse inclusion, $A_j\subseteq A$ for every $j$. The smallest-closed-superset property implies $\overline{A_j}\subseteq\overline{A}$, since $\overline{A}$ is a closed set containing $A_j$. Taking the union over all $j$ gives \[ \bigcup_{j=1}^{r}\overline{A_j}\subseteq\overline{A}. \] The two inclusions prove the equality. $\square$

Worked Example: Closing a Union with a Missing Point

Let $A_1=(1,4)$ and $A_2=(4,7)$. The closure of the first interval is $[1,4]$, and the closure of the second is $[4,7]$. Applying the Closure of a Finite Union Theorem gives

\overline{(1,4)\cup(4,7)} =[1,4]\cup[4,7] =[1,7].

The point $4$ is absent from the original union, but it belongs to the closure of each interval: points of each interval occur arbitrarily close to $4$. The two closed pieces meet at $4$, so their union is the full closed interval $[1,7]$. In this calculation, closing the two pieces separately makes the endpoint behavior explicit.

Finiteness is essential in this closure identity. It is not true for arbitrary unions in general: taking the closure of an infinite union can add a point approached by points drawn from more and more different pieces, even if that point is not in the closure of any individual piece. The next result explains why such an effect cannot arise from only finitely many pieces.

Limit Points of a Finite Union

Recall that $x$ is a limit point of a set $E$ if every punctured neighborhood of $x$ meets $E$. The Sequential Characterization of Limit Points, proved earlier in this course, says equivalently that there is a sequence of points of $E\setminus\{x\}$ converging to $x$. For finite unions, the limit-point set can be determined piece by piece.

Theorem (Limit Points of a Finite Union): If $A_1,\ldots,A_r\subseteq\mathbb{R}$, where $r$ is a positive integer, then the set of limit points of $\bigcup_{j=1}^{r}A_j$ is the union of the limit-point sets of $A_1,\ldots,A_r$.

Proof. First suppose $x$ is a limit point of $\bigcup_{j=1}^{r}A_j$. By the Sequential Characterization of Limit Points, there is a sequence $(x_n)$ with $x_n\in\bigcup_{j=1}^{r}A_j$, $x_n\neq x$, and $x_n\to x$. For each $n$, choose one index $j_n\in\{1,\ldots,r\}$ such that $x_n\in A_{j_n}$. Since there are only finitely many possible indices, at least one index $j$ occurs for infinitely many $n$. Restrict to those indices in increasing order. The resulting subsequence lies in $A_j$, still avoids $x$, and still converges to $x$. The sequential characterization therefore shows that $x$ is a limit point of $A_j$.

Conversely, suppose $x$ is a limit point of $A_j$ for some $j$. Every punctured neighborhood of $x$ then meets $A_j$. Since $A_j\subseteq\bigcup_{i=1}^{r}A_i$, that neighborhood also meets the union. Thus $x$ is a limit point of the union. Both inclusions have been proved, so the limit-point set of the union is exactly the union of the component limit-point sets. $\square$

Worked Example: Two Sequences with Different Limit Points

For positive integers $n$, define

A_1=\left\{3+\frac{1}{n}:n\geq1\right\}, \qquad A_2=\left\{-2-\frac{1}{n}:n\geq1\right\}.

The sequence $3+1/n$ converges to $3$, and none of its terms equals $3$, so $3$ is a limit point of $A_1$. There are no other limit points of $A_1$: any convergent sequence of distinct elements from this set has indices tending to infinity, and its terms therefore converge to $3$. Likewise, $-2-1/n\to-2$, with no term equal to $-2$, and the only limit point of $A_2$ is $-2$.

The Limit Points of a Finite Union Theorem now shows that the limit points of $A_1\cup A_2$ are exactly $3$ and $-2$. In particular, the two sequences do not create an additional limit point between their two clusters. Any sequence drawn from their union that converges must have a subsequence drawn from one of the two pieces, and that subsequence has the corresponding piece's limit point.

Why Infinite Unions Are Different

For a finite union, the same component must contain infinitely many terms of any sequence drawn from the union. With infinitely many components, the terms can come from a different component at each stage. Consequently, a convergent sequence in the union need not have infinitely many terms in any one fixed component. This is the point at which the finite proofs above stop working.

Worked Example: A Countable Union of Closed Sets That Is Not Closed

For each positive integer $n$, let $F_n=\{1/n\}$. A singleton is closed in $\mathbb{R}$, so every $F_n$ is closed. Their union is

\bigcup_{n=1}^{\infty}F_n = \left\{\frac{1}{n}:n\geq1\right\}.

The sequence $1/n$ consists entirely of points in this union and converges to $0$. But $0$ is not in the union, since $1/n$ is positive for every positive integer $n$. The Sequential Characterization of Closed Sets therefore shows that the union is not closed. Each individual set contributes only one point; no fixed singleton contains infinitely many terms of the sequence. This example demonstrates why the finite-union theorem cannot be extended to arbitrary unions.

The limitation concerns unions, not all operations on closed sets. The Arbitrary Intersections of Closed Sets Theorem says that intersections of any family of closed sets remain closed. For unions, the safe rule is the finite one: a finite union of closed sets is closed, while an infinite union requires a separate argument about the particular sets involved.

Decompose the set into finitely many pieces.
Identify familiar closed pieces when proving closedness, or simpler pieces when calculating closure or limit points.

Use the appropriate finite-union result.
Closed pieces have a closed union; closure and limit points of a finite union can be computed piece by piece.

Check whether the family is actually finite.
An infinite union may fail to be closed, so do not apply a finite-union conclusion without verifying the number of pieces.

A common pitfall is to reason that every point of a union belongs to one closed component and therefore the union must be closed. Closedness is also about limits of sequences, and a sequence can move among components. With finitely many components, one component must recur infinitely often; with infinitely many, that need not happen. This distinction explains both the theorem and its boundary.

Check Your Understanding

Use the finite-union results and examples to answer the following questions.

What hypothesis on the number of closed sets is required by the Finite Unions of Closed Sets Theorem?
State the formula for the closure of a finite union.
Why does a sequence in a finite union have a subsequence lying in one fixed member of the union?
What are the limit points of $\{3+1/n:n\geq1\}\cup\{-2-1/n:n\geq1\}$?
Why is $\bigcup_{n=1}^{\infty}\{1/n\}$ not closed?
Which feature of an infinite union prevents the finite-union theorem from applying directly?

What You'll Learn